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Contributions to Algebra and Geometry Volume 43 (2002), No. 2, 303-324.

### Part II: Maps on Points and/or Lines

Eline Govaert Hendrik Van Maldeghem

University of Ghent, Pure Mathematics and Computer Algebra Galglaan 2, 9000 Gent, Belgium

e-mail: egovaert@@cage.rug.ac.be; hvm@@cage.rug.ac.be

Abstract. In this paper, we characterize isomorphisms of generalized polygons (in particular automorphisms) by maps on points and/or lines which preserve a certain fixed distance. In Part I, we considered maps on flags. Exceptions give rise to interesting properties, which on their turn have some nice applications.

MSC 2000: 51E12

Keywords: generalized polygons, distance-preserving maps

1. Introduction

For an extensive introduction to the problem, we refer to Part I of this paper. Let us briefly describe the situation we are dealing with.

For the purpose of this paper, a generalized n-gon, n ≥3, is a thick geometry such that every two elements are contained in some ordinary n-gon, and no ordinary k-gons exist for k < n.

Given two generalized n-gons, and a map from either the point set or the point set and the line set of the first polygon to the point set, or the point set and the line set, respectively, of the second polygon, preserving the set of pairs of elements at a certain fixed distancei≤n, we investigate when this map can be extended to an isomorphism (“preserving” means that elements of the first polygon are at distanceiif and only if their images in the second polygon

The first author is a research assistant of the FWO, theFund for Scientific Research – Flanders (Belgium)

0138-4821/93 \$ 2.50 c 2002 Heldermann Verlag

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are at distancei). In the first part of this paper, we dealt with the same problem considering maps on the set of flags. More precisely, we showed (denoting theorderof a polygon by (s, t) if there are s+ 1 points on every line, and t+ 1 lines through every point):

Theorem 1. Letand0 be two generalized m-gons, m≥2, let r be an integer satisfying 1≤ r ≤ m, and let α be a surjective map from the set of flags ofonto the set of flags of

0. Furthermore, suppose that the orders ofand0 either both contain 2, or both do not contain 2. If for every two flags f, g of ∆, we have δ(f, g) = r if and only if δ(fα, gα) = r, then α extends to an (anti)isomorphism fromto0, except possibly whenand0 are both isomorphic to the unique generalized quadrangle of order (2,2) and r= 3.

In [7], we also gave an explicit counterexample for the quadrangle of order (2,2).

In this part, we will show:

Theorem 2.

Let Γ and Γ0 be two generalized n-gons, n ≥ 2, let i be an even integer satisfying 1≤i≤n−1, and let α be a surjective map from the point set of Γ onto the point set of Γ0. Furthermore, suppose that the orders of Γ and Γ0 either both contain 2, or both do not contain 2. If for every two points a, b of Γ, we have δ(a, b) = i if and only if δ(aα, bα) =i, then α extends to an isomorphism from Γ to Γ0.

LetΓ andΓ0 be two generalizedn-gons,n ≥2, let i be an odd integer satisfying1≤i≤ n−1, and letα be a surjective map from the point set ofΓ onto the point set of Γ0, and from the line set of Γ onto the line set ofΓ0. Furthermore, suppose that the orders of Γ andΓ0 either both contain2, or both do not contain2. If for every point-line pair{a, b}

of Γ, we have δ(a, b) = i if and only if δ(aα, bα) = i, then α extends to an isomorphism from Γ to Γ0.

As already mentioned in Part I, there do exist counterexamples for the case n = i, and we will construct some in Section 3, where we also prove two little applications. Also, the condition i 6=n can be deleted for n = 3,4, of course, in a trivial way. For finite polygons, the condition i 6= n is only necessary if n = 6 and the order (s, t) of Γ satisfies s = t. For Moufang polygons, the condition i 6= n can be removed if Γ is not isomorphic to the split Cayley hexagon H(K) over some field K (this is the hexagon related to the group G2(K)).

We will prove these statements in Section 3.

To close this section, we repeat the notation that we use throughout the two parts of this paper. Let Γ be a generalizedn-gon. For any point or linex, and any integeri≤n, we denote by Γi(x) the set of elements of Γ at distance i from x, and we denote by Γ6=i(x) the set of elements of Γ not at distanceifromx. Ifκis a set of integers, then Γκ(x) is the set of elements y of Γ satisfying δ(x, y) ∈ κ. If two elements are at distance n, then we say that they are opposite. Non-opposite elementsxandyhave a unique shortest chain (x=x0, x1, . . . , xk =y) of length k =δ(x, y) joining them. We denote that chain by [x, y], and we set x1 = projxy (and hence xk−1 = projyx). When it suits us, we consider a chain as a set so that we can take intersections of chains. For instance, if [x, z] = (x = x0, x1, . . . , xi, x0i+1, . . . , x0` = z), with no x0j equal to any xj0, 0< i < j ≤k and i < j0 ≤ `, then [x, y]∩[x, z] = [x, xi]. If for two non-opposite elements x, y the distance δ(x, y) is even, then there is a unique element

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z at distance δ(x, y)/2 from both x and y; we denote z =x1y, or, if x and y are points at distance 2 from each other, then we also write xy:=x1y.

In the next section, we will prove Theorem 2. In Section 3, we produce some counterex- amples and prove some applications.

2. Proof of Theorem 2

As in the proof of Theorem 1, we again see that α is necessarily bijective.

We again prove the assertion in several steps, the general idea being to show that collinear- ity of points is preserved (then Lemma 1.3.14 of [9] gives the result).

Throughout we putTa,b:= Γi(a)∩Γi(b), for points a, bof Γ.

2.1. Case i < n−12 , with i even

This is the easy case. Putλ :={2i+ 2, . . . , n} 6=∅ and κ:={0,1, . . . , i−2}. Then one can easily check that for two arbitrary distinct points a, b of Γ, we have Ta,b = ∅ if and only if δ(a, b)∈ λ. Also, it is easily verified that, if δ(a, b) ∈/ λ∪ {i}, then Γi(a)∩Γλ(b) = ∅ if and only if δ(a, b)∈κ. Now clearly, if δ(a, b)∈κ, then

Γκ(a)⊆Γκ(b)∪Γi(b)⇐⇒δ(a, b) = 2.

Hence α preserves collinearity and the assertion follows.

2.2. Case i = n−12 , with i even

Here, Ta,b is never empty, for all pointsa, b of Γ.

First supposes >2. Let a, bbe arbitrary points of Γ. Then clearly |Ta,b|= 1 if and only if δ(a, b) = 2i. So we can distinguish distance 2i. Also, it is clear that Γi(a)∩Γ2i(b) = ∅ if and only if δ(a, b)< i. Now one proceeds as in case 2.1.

Next suppose s = 2. Then |Ta,b| = 1 if and only if δ(a, b) ∈ {2i−2,2i}. Also, Γi(a)∩

Γ{2i−2,2i}(b) =∅ if and only ifδ(a, b)< i−2. Then similarly as before, ifδ(a, b)< i−2, then

Γi(a)∩Γ<i−2(b) = ∅ ⇐⇒δ(a, b) = 2.

So again, α preserves collinearity.

The next two cases have some overlap. In particular, if i = (n+ 1)/2, then two proofs apply.

2.3. Case i ∈ {n2 + 1,n+12 }, i even and n >6

LetSbe the set of pairs of distinct points (a, b) such thatδ(a, b)6=iand the setTa,b contains at least two points at distance i from each other. We claim that a pair (a, b) belongs to S if and only if δ(a, b)< i. Suppose first that 06=δ(a, b) =k, k < iand put m=a1b. Consider a point c at distancei−k/2 from m such that projma 6= w := projmc6= projmb (note that δ(c, w) = i−k/2−1 > 0). Let v be the element of [c, w] at distance i/2 from c (such an element exists since i/2≤i−k/2−1). Consider a pointc0 at distance i/2 from v such that

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projvm 6= projvc0 6= projvc. The points c and c0 are both points of Ta,b and lie at distance i from each other.

Now let δ(a, b) = k > i and suppose by way of contradiction that c, c0 ∈ Ta,b with δ(c, c0) = i. If projac6= projac0, then we have a path of length 2ibetweencandc0 containinga.

This implies that δ(c, c0)≥2n−2i > i, a contradiction. Suppose now that projac= projac0. Define v as [a, c] ∩[a, c0] = [a, v]. If we put δ(a, v) = j, then there is a path of length

` = 2i−2j ≤ n between c and c0. Now ` = i implies j = i/2. If the path [b, c] does not contain v, there arises a circuit of length at most 3i < 2n, a contradiction (remembering n > 6). But if v ∈ [b, c], there arises a path between a and b of length at most i, the final contradiction. Our claim is proved.

Putκ={1,2, . . . , i−2}. Then (a, b)∈S if and only if δ(a, b)∈κ.

Now two distinct points a and b are collinear if and only if δ(a, b) ∈ κ and Γκ(a) ⊆ Γκ(b) ∪ Γi(b). Indeed, let δ(a, b) = k, 2 < k < i. Then k = i − j, 0 < j < i− 2.

Consider a point cat distance j+ 2 froma such that projac6= projab. Thenδ(a, c)∈κ, but δ(b, c) = i+ 2. Ifδ(a, b) = 2, then the triangle inequality shows the assertion. Hence we can distinguish distance 2 and so α preserves collinearity of points.

2.4. Case n+12 ≤i < n−2, with i odd if i = n2 + 1

LetS be the set of pairs of points (a, b) for which there exists a point c, a6=c6=b, such that Ta,b ⊆ Γi(c). Note that Ta,b is never empty because n/2 < i. We claim that the pair (a, b) with δ(a, b) = k belongs to S if and only if 2 < k < 2(n−i). Note that there are always even numbersk satisfying these inequalities becausei < n−2. So let a, bbe points of Γ with δ(a, b) = k and let m be an element at distance k/2 from both a and b (m is unique if a is not opposite b).

We first show that k < 2(n −i) if and only if every element y of Ta,b lies at distance i−k/2 from m with projma 6= projmy 6= projmb. Suppose first that k ≥ 2(n−i). Since k+ 2i ≥ 2n, it is possible to find an element y of Ta,b such that projay 6= projab. Clearly, δ(y, m) 6= i−k/2. Now suppose k < 2(n−i) and let y ∈ Ta,b. Let j be the length of the path [a, b]∩[a, y]. Then there is a path of length ` =k−2j+i between b and y. If ` ≤n, then ` = i and so necessarily j = k/2. Hence y is an element as claimed. If ` > n, then δ(b, y)≥2n−` > i, a contradiction.

From this, it is clear that all pairs (a, b) with 2 < δ(a, b) = k < 2(n−i) belong to S (indeed, choose the point c on the line projab, at distance k from b).

We now show that

(∗) every point cat distanceifrom every element ofTa,bhas to lie at distancek/2 fromm, and projma= projmc or projmb= projmc.

Suppose cis a point at distance i from every element of Ta,b. Consider the set T0 ={x∈Ta,b|δ(x, m) = i−k/2 and projma6= projmx6= projmb}.

Then we may assume that T0 contains at least two elements yand y0 at distance 2 from each other (this is clear either if i > n/2 + 1, or if n is odd, or if t > 2 in case i = n/2 + 1 and i is odd; if t = 2 and i = n/2 + 1 with i odd, then we consider the dual of Γ; finally the

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case i = n/2 + 1 with i even is not included in our assumptions, see Subsection 2.3). Put w = y1y0. Then δ(c, w) = i−1. Put γ = [c, w]. We show that γ contains m. Suppose by way of contradiction that this is not true. Define the element z as [w, c]∩[w, m] = [w, z].

Putγ0 = [z, c]. An element y00 of T0 either lying onγ0 (if δ(c, m)≥i−k/2) or such that the path [y00, m] contains γ0 (otherwise), clearly does not lie at distanceifromc, a contradiction.

So the point c has to lie at distance k/2 from m. But if projma 6= projmc 6= projmb, then similarly we can find an element of T0 not at distance i fromc, which shows (∗).

Letk = 2 ork ≥2(n−i). We show that if a pointclies at distanceifrom every element of Ta,b, thenc∈ {a, b}. Ifk= 2, then this follows immediately from (∗). So supposek≥2(n−i) and let c be such a point. We may assume that, if δ(m, c) 6= n, then projma = projmc. If δ(a, c)6=n, then we define the element z as [m, c]∩[m, a] = [m, z]; otherwise we define z as [projma, c]∩[projma, a] = [projma, z]. Note that δ(c, z) =δ(a, z) =:`. Put j =i−n+k/2.

It is easy to check that for an elementv of the path [a,projma], the following property holds.

(∗∗) There exists y∈Ta,b such that [a, y]∩[a, m] = [a, v] if and only if δ(a, v)≤j.

It follows from (∗∗) that`≤j (indeed, if y∈Ta,b is such thatm /∈[a, y], then the path [c, y]

is longer than the path [a, y]).

But similarly, if `≤j, then an elementy ∈Ta,b such that projzc∈ [a, y], does not lie at distancei from c, a contradiction.

This shows our claim. Note that, if 2< k <2(n−i) and if c is a point of Γ at distance k/2 from m with projmc∈ {projma,projmb}, then automaticallyTa,b⊆Γi(c).

Now defineS0 ={(a, c)| ∃b ∈ P such that Ta,b ⊆Γi(c)}. From the previous paragraph it is clear that S0 \S is precisely the set of all pairs of collinear points.

Hence α preserves collinearity.

2.5. Case i = n−2 2.5.1. Case n = 6

Let C be the set of pairs of points (a, b), δ(a, b) 6= 4, such that for every point y in Ta,b, there exists a point y0 in Ta,b, y0 6=y and δ(y, y0) 6= 4, with the property that Γ4(y)∩Ta,b = Γ4(y0)∩Ta,b. Clearly, C contains all pairs of collinear points. Suppose now that (a, b) ∈ C with δ(a, b) = 6. We look for a contradiction. If x is a point of Ta,b, then either x lies on a line at distance 3 from both a and b, or x is a point at distance 3 from a line A through a and from a line B through b, with A opposite B. Then one can check that for a point y of Ta,b on a line at distance 3 from both aand b, there does not exist a pointy0 6=y inTa,b such that Γ4(y)∩Ta,b = Γ4(y0)∩Ta,b. So the set C is the set of pairs of collinear points and the theorem follows.

2.5.2. Case n >6 Step 1: the set Sa,b

For two pointsa and b, we define

Sa,b ={x∈ P |Γn−2(x)∩Ta,b=∅}.

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We claim the following:

(i) If δ(a, b) = 2 ands ≥3, then Sa,b = (Γ2(a)∪Γ2(b))\Γ1(ab). If δ(a, b) = 2 ands = 2, then Sa,b= (Γ2(a)∪Γ2(b))\ {a, b}.

(ii) If δ(a, b) = 4, then {a1b} ⊆ Sa,b = {a1b} ∪[Γ2(a1b)∩Γ4(a)∩Γ4(b)]. If t ≥ 3, then Sa,b = {a1b}. If k := δ(a, b) 6∈ {2,4, n}, then every x ∈ Sa,b lies at distance k/2 from a1b =: m with projma 6= projmx 6= projmb. If moreover s > 2 and k ≡ 2 mod 4, or t >2 andk ≡0 mod 4, then Sa,b =∅. Finally, ifδ(a, b) = n, then let γ be an arbitrary path of lengthnjoiningaandb, letmbe the middle element ofγ and putva= projma, vb = projmb. Then

Sa,b ⊆ (Γn/2(m)∩Γn/2+1(va)∩Γn/2+1(vb)) [

n/2+1(va)∩Γn/2+2(m)∩Γn(a)) [

n/2+1(vb)∩Γn/2+2(m)∩Γn(b)).

If moreover s >2 and n≡2 mod 4, or t >2 and n≡0 mod 4, then Sa,b ⊆ (Γn/2+1(va)∩Γn/2+2(m)∩Γn(a))

[

n/2+1(vb)∩Γn/2+2(m)∩Γn(b)).

We prove these claims.

(i) Suppose δ(a, b) = 2. Clearly, every point collinear with a or b, not on the line ab, belongs toSa,b. Also, if s= 2, then the unique point of abdifferent from a and b is an element ofSa,b. Let xbe an arbitrary point inSa,b. Put j =δ(x, a). If j =s = 2, then there is nothing to prove, so we may assume (j, s) 6= (2,2). Suppose first there exists a j-path γ between a and x containing ab, but not the point b. Let v be the element on γ at distance j/2 from a, and consider an element y at distance n−2−j/2 from v such that projva 6= projvy 6= projvx. Note that such an element v exists because (j, s) 6= (2,2). Then y lies at distance n−2 from a, b and x, a contradiction. So we can assume that projabx = a. If j = 2, then again, there is nothing to prove. So we may assume 2 < j < n (the case j = n is contained in the previous case, or can be obtained from the present case by interchanging the roles of a and b). Let v be an element at distance n−j −1 from the line ab such that a 6= projabv 6= b. Note that v and x are opposite and δ(a, v) = n −j. Consider an element v0 incident with v, different from projva, and let v00 be the projection of x onto v0. Let w be the element of [x, v00] at distance j/2−2 from v00. An element y at distance j/2−2 from w such that projwx6= projwy6= projwv00 lies at distancen−2 froma,b and x, a contradiction.

Claim (i) is proved.

(ii) We proceed by induction on the distance k between a and b, the case k = 2 being claim (i) above. Supposeδ(a, b) =k > 2 and let m be an element at distancek/2 from botha and b. Note that, if δ(a, b) = 4, the point a1b indeed belongs toSa,b. Let now x be an arbitrary element of Sa,b and put `=δ(x, m).

Suppose first that, if ` 6= n, projma 6= projmx 6= projmb. Then we have the following possibilities :

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1. Suppose ` < k/2. Then δ(a, x)< k and we apply the induction hypothesis. Since b ∈ Sa,x 6= ∅ and m 6= a1x, we have δ(a, x) ∈ {2,4}. Hence either δ(a, b) = 4 and x = a1b (which is a possibility mentioned in (ii)), or δ(a, b) = 6 and x lies onm, or δ(a, b) = 8 and x =m. But in these last two cases, the “position” of b contradicts the induction hypothesis.

2. Suppose` ≥k/2. Letγ00 be an`-path betweenm andxcontaining neither projma nor projmb. Putγ0 = [a, m]∪γ00. Letwbe the element onγ0 at distance (`+k/2)/2 from both x and a. If ` =k/2 (and hence w = m) and either k ≡ 2 mod 4 and s = 2, or k ≡ 0 mod 4 and t = 2, then there is nothing to prove. Otherwise, there exists an element y of Γ at distance n−2−(k/2 +`)/2 from w such that projwa 6= projwy 6= projwx and projwb 6= projwy. Now y lies at distance n−2 froma, b and x, a contradiction.

Let nowxbe a point ofSa,bat distance`fromm, 0< ` < n, for which projmx= projma.

Let [a, m]∩[x, m] = [v, m], and put i0 =δ(v, a). We have the following possibilities : 1. Suppose `≤k/2 or `=k/2 + 2 and i0 < k/2−1. Againδ(a, x)< k and applying

the induction hypothesis, we obtain a contradiction as in Case 1 above.

2. Suppose n > ` > k/2 + 2. Leth be an element at distancen−2 fromxsuch that projma6= projmh6= projmb andδ(m, h) =n−`+ 2. Letj =n−2−δ(h, m)−k/2.

Leth0 be the element on the (n−2)-path between x and h at distance j/2 from h. An elementy at distancej/2 fromh0 such that projh0x6= projh0y 6= projh0hlies at distancen−2 from a, b and x, a contradiction.

3. Suppose ` =k/2 + 2, i0 =k/2−1 andk < n−1. Then δ(b, x) =k+ 2 andv lies at distancek/2 + 1 from bothb andx. Let Σ be an apartment containingx,b and v, and letv0 be the element in Σ opposite v. Letw= projva, w0 = projv0wand d the length of the path [w, a]∩[w, w0]. Note thatd ≤k/2−2. For an elementynot opposite w0, let wy00 be the element such that [w, w0]∩[y, w0] = [w00y, w0]. Consider now an elementy such that δ(w00y, w0) =k/2−d−2 andδ(w00y, y) =d. Thenylies at distancen−2 from a, b and x, a contradiction.

4. If `=k/2 + 2,i0 =n/2−1 and k =n, there is nothing to prove.

5. Suppose finally` =k/2 + 2,i0 =k/2−1 and k =n−1. Then δ(b, x) =n−1. Let b0 and x0 be the elements of the path [b, x] at distance (n−1)/2−1 from b and x, respectively. Since a ∈Sb,x, either δ(a, b0) = (n+ 1)/2 or δ(a, x0) = (n+ 1)/2 (this is what we proved up to now for the “position” of a point ofSb,x). But since we obtain a path betweena and b0 (x0) of lengthd= (3n−5)/2 (passing through projma), the triangle inequality implies δ(a, b0), δ(a, x0) ≥ 2n−d > (n + 1)/2, a contradiction.

This completes the proof of our claims.

Step 2: the sets O and O

Ifs≥3 andt ≥3, letO be the set of pairs of distinct points (a, b) such that|Sa,b|>1. Then O contains only pairs of collinear points and pairs of opposite points, and all pairs of collinear

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points are included inO. Note that, if n is odd, there are no pairs of opposite points, which concludes the proof in this case.

Let now s= 2 or t= 2 (so n is even). For a point c∈Sa,b, we define the set Ca,b;c ={c0 ∈Sa,b|Sc,c0 ∩ {a, b} 6=∅}.

Now let O be the set of pairs of points (a, b), δ(a, b) 6= n − 2 for which |Sa,b| > 1 and

|Ca,b;c| > 1, ∀c ∈ Sa,b. Note that if a and b are collinear, the pair (a, b) always belongs toO. Clearly, no pair of points at mutual distance 4 belongs to O. Now consider two points a and b at distance k, 4 < k < n −2. We show that such a pair (a, b) does not belong to O. Putm =a1b and let x be a fixed point of Sa,b. Let x0 be an element of Sa,b different from x. Since s = 2 or t = 2, we have projmx = projmx0, so δ(x, x0) ≤ k −2. But now δ(a, x1x0) = δ(b, x1x0)≥k/2 + 1> δ(x, x0)/2, so neitheranorb belongs toSx,x0. This shows thatO contains only pairs of collinear or opposite points, and all pairs of collinear points are included in O.

Let O be the set of pairs of points (a, b) satisfying δ(a, b) 6= n−2, (a, b) 6∈ O and such that there exist a point c∈Sa,b for which (a, c) and (b, c) both belong to O. Then clearly, O contains all pairs (a, b) of points at mutual distance 4 (indeed, consider the point c=a1b), and also some pairs of opposite points (possibly none, or all).

Step 3: the set O0 Suppose first s≥3.

Let O0 be the subset of O of pairs (a, b) for which there exist points c and c0 such that the following conditions hold :

(i) Ta,b ⊆Γn−2(c)∪Γn−2(c0),Tc,c0 ⊆Γn−2(a)∪Γn−2(b);

(ii) (c, c0),(a, c),(a, c0),(b, c),(b, c0)∈O;

(iii) (c, y),(c0, y)6∈O, ∀y∈Ta,b.

We claim that O0 is the set of pairs of collinear points. A pair (a, b) of collinear points always belongs to O0. Indeed, here, we can choose c and c0 on the line ab, different from a and b (Condition (iii) is satisfied because n 6= 6). So let δ(a, b) = n and suppose by way of contradiction that we have two points c and c0 with the above properties. Let m ∈ Γn/2(a)∩ Γn/2(b). For an element x at distance j from m, 0 ≤ j ≤ n/2− 3, such that projma6= projmx6= projmb, define the following set:

Tx ={y∈Ta,b|δ(x, y) =n/2−2−j,projxa6= projxy6= projxb}.

Note thatTx ⊆Ta,b. We first prove that for any set Tx,

(3) there does not exist a point v ∈ {c, c0} such thatTx⊆Γn−2(v).

Put M = Γn/2(a)∩Γn/2(b). Suppose Tm ⊆ Γn−2(v), with v ∈ {c, c0}. It is easy to see (see for instance Step 4 of Subsection 3.4 of [7]) that δ(v, m) = n/2 and projma = projmv or projmb= projmv. Suppose projma= projmv. But then δ(a, v)≤n−2, so δ(a, v) = 2 (since (a, v)∈O) andvis a point at distancen/2 frommlying on the lineL= projam. This implies that for an arbitrary point m0 of M, m0 6=m, Tm0 ∩Γn−2(v) =∅ (note that Tm∩Tm0 =∅),

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so Tm0 ⊆ Γn−2(v0), with {v, v0} = {c, c0}. We obtain a contradiction by considering a third element of M.

Letx be an element at distance j = 1 from m such that projma6=x6= projmb. Suppose Tx ⊆Γn−2(v), withv ∈ {c, c0}. Then again it is easy to show that δ(v, x) = δ(x, a) = n/2 + 1 and projxv = projxa =m. If projma= projmv or projmb = projmv, then we are back in the previous case, which led to a contradiction, so suppose projma 6= projmv 6= projmb. Consider the n-path between aand v that containsm. Then we can find a pointyof Ta,b on this path that is collinear with v, in contradiction with condition (iii). Note that thus no element of {c, c0} lies at distance n/2 from m.

We now proceed by induction on the distance j between x and m. Let j > 1. Consider an element xat distancej fromm such that projma6= projmx6= projmb. Suppose by way of contradiction that Tx ⊆Γn−2(v), with v ∈ {c, c0}. Letx0 = projxm. Then it is again easy to show that δ(v, x) =δ(a, x) =n/2 +j and projxv =x0. Remark that projx0a6= projx0v, since otherwiseTx0 ⊆Γn−2(v) (sinceδ(v, x0) =n/2 +j−1 andδ(v, w) =n/2−j−1, withw∈Tx0), in contradiction with the induction hypothesis. Suppose first that in the case j = 2, t≥3 or n≡2 mod 4. Consider now an elementz incident with the elementw= projx0a, but different from projwa, from projwb and from x0 (such an element exists, because of the restrictions above). But then we have δ(v, w0) = n, for every element w0 of Tz, so Tz ⊆ Γn−2(v0), with {c, c0}={v, v0}, a contradiction with the induction hypothesis.

Now let j = 2, t = 2 and n ≡ 0 mod 4. Let L be the line mx and put w = projLv.

Then Tw ⊆Γn−2(v0), with {v, v0} ={c, c0}, sov0 is a point at distance n/2 + 2 from m with δ(m, v0) 6= n/2, and projLv0 6= w. Now consider the point on [m, b] at distance n/2−4 from m. This is a point of Tc,c0, but it does not lie at distance n−2 from a, nor from b, in contradiction with condition (i). This completes the proof of (3).

Consider now a lineLat distance j =n/2−3 from m, such that projma6= projmL6= projmb.

The points onLdifferent from the projection ofm ontoLare points ofTL. By (3), we know that TL 6⊆ Γn−2(v), for v ∈ {c, c0}. Since s ≥ 3, TL contains at least 3 points, so we may suppose that at least two points of them are contained in Γn−2(v), with v ∈ {c, c0}. This implies that v is at distancen−3 fromL, so at distancen−4 from a unique pointxof L. If x= projLa, thenTL ⊆Γn−2(v), a contradiction, so we can assume thatx6= projLa. Let first n6= 8 or t≥3. Then consider a line L0 incident with projLa,L0 6=L, at distance n−3 from botha and b (such a line always exists because of our assumptions). NowTL0 ∩Γn−2(v) = ∅ (because all points ofTL0 lie oppositev), soTL0 is contained in Γn−2(v0), with{v, v0}={c, c0}, the final contradiction.

Let now n = 8 and t = 2. Then δ(v0, x) = 6, with {v, v0} = {c, c0}, TL 6⊆ Γ6(v0) and δ(v, v0)∈ {2,8}. Now for each potential v0, it is possible to construct a point of Tv,v0 not at distancen−2 from a nor from b, a contradiction with condition (i). For example, let us do in detail the case δ(v, v0) = 2. Since v does not lie at distance 6 from a or b, we know that δ(a, vv0) = δ(b, vv0) = 7, v0 6= projvv0a 6= v and v0 6= projvv0b 6= v. Also projvv0a 6= projvv0b, since otherwise we would obtain a point ofTa,b not at distance 6 fromv nor fromv0. Now let N be the line at distance 3 fromb and at distance 4 fromvv0. Then the points ofN different from projNv are points of Tv,v06(b), but not all these points lie at distance 6 from a, a contradiction.

So in the caseδ(a, b) = n, pointscand c0 with the above properties cannot exist, and the

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proof is finished for s≥3.

Suppose nows = 2.

Let first n = 8. Note that t ≥ 4. Let O0 be the subset of O of pairs of points (a, b) for which there exist a pointc∈Sa,band pointsc0,c00 belonging toCa,b;c such thatSc,c0∩ {a, b}= Sc,c00∩ {a, b}=Sc0,c00∩ {a, b}={a}. We claim thatO0 is the set of all pairs of collinear points.

Letδ(a, b) = 2. Then considering three pointsc, c0 and c00 on three different lines through a, different from ab, shows that (a, b) ∈ O0. Now let δ(a, b) = 8, and suppose (a, b) ∈ O0. Let m be any point at distance 4 from both a and b. Put a0 = projma and b0 = projmb. Suppose that the pointcmentioned in the property above lies at distance 5 from the linea0 (which is allowed by Step 1 above). Then it is easy to see that c0 and c00 both have to lie at distance 5 fromb0, which contradicts Sc0,c00∩ {a, b}={a}. Similarly ifc lies at distance 5 from b0. Suppose from now on n >8. Let O0 be the subset of O of pairs (a, b) for which

(i) ∃!x1 ∈Sa,b:∀x0 ∈Sa,b,|Sx1,x0 ∩ {a, b}|= 1,

(ii) there exist pointsc and c0 such that Ta,b ⊆Γn−2(c)∪Γn−2(c0), (iii) (x1, v)∈O, with v an element of the set {a, b, c, c0},

(iv) (a, c),(a, c0),(b, c),(b, c0),(c, c0)∈O, (v) (c, y),(c0, y),(x1, y)6∈O, ∀y∈Ta,b, (vi) x1 =Sa,c∩Sa,c0 ∩Sb,c∩Sb,c0 ∩Sc,c0.

We claim that O0 is the set of pairs of points at distance 2.

The claim is clear for two collinear points a and b. Indeed, let x1 be the unique point on ab, different from a and b, and c and c0 points on two different lines (different from ab) through the point x1 (Condition (v) in the definition of O0 does not hold if n = 8, which is the reason we treated the octagons before).

So let δ(a, b) =n and suppose by way of contradiction that we have two points c and c0 with the above properties. Let γ be a fixed n-path between a and b, and define m, a0 and b0 as above in the case n = 8.

For an element x at distance j from m, 0 ≤j ≤n/2−3, such that projma 6= projmx6=

projmb, we define the sets Tx in the same way as for the case s ≥ 3. We again first prove thatTx6⊆Γn−2(c), for all sets Tx. Now, everything can be copied from the cases ≥3, except when j = 0 or j = 2.

(j = 0) Suppose Tm ⊆ Γn−2(v), with v ∈ {c, c0}. Then we know that δ(v, m) = n/2 and we may assume projmv = projma. This implies that δ(a, v) ≤n−2, so δ(a, v) = 4 (see Condition (iv)). Then x1 is the point of the path γ collinear with a (since Sa,v contains only the element a1v in this case), so δ(b, x1) = n−2, which contradicts the fact that (x1, b)∈O.

(j = 2) Note that this case is a problem only when n≡2 mod 4, so we may assume thatm is a line. Suppose TL ⊆Γn−2(v), for a line L concurrent withm, at distance n/2 + 2 from a and b, and for a point v ∈ {c, c0}. Then δ(v, L) = δ(a, L) = n/2 + 2 and projLa = projLv. We may again assume that v does not lie at distance n/2 from m.

By Step 1, there are essentially two possibilities for x1.

First, suppose the point x1 lies at distance n/2 + 2 from m and at distance n/2 + 1 from a0. Then there arises an n-path γ0 between a and x1 sharing the path [a, a0]

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withγ. Leta00be the projection of x1 ontoa0. Sincea00 is a line at distancen/2 from both a and x1, and v ∈Sx1,a, either the distance between v and a00 is n/2 (which is not true), or the distance between v and a00 is n/2 + 2, which is again impossible.

Secondly,x1 cannot lie at distancen/2 fromm, since this would contradict condition (v) (x1 would be collinear with a point of Ta,b).

Hence (3) is proved in this case.

Now we have to find an alternative argument for the last paragraph of the general case, since we relied there on the fact that a line contains at least 4 points. We keep the same notation of that paragraph. Now the only possibility (to rule out) that we have not considered yet (because it does not occur in the general case) is the case that c and c0 both lie at distance n−2 from different points u and u0 on L, δ(c, L) = δ(c0, L) = n−1 and u and u0 different from the projection w of a onto L.

Suppose first n > 10 (otherwise some of the notations introduced below don’t make sense). Put L0 = projwa and l0 = projL0a. Suppose the unique point z on L0 at distance n−2 from cis not l0. Then consider a line K through z, different from L0 and from projzc.

Because c is at distance n from all the points of K, different from z (which are elements of Ta,b) we can conclude that TK ⊆ Γn−2(c0), a contradiction to (3). So [c, L0] contains l0. Define the element p as [l0, m]∩[l0, c] = [l0, p]. Suppose p6=m and let j =δ(l0, p). Consider the element z0 on [c, p] at distancej+ 3 from p. Note that z0 is a line at distance n−3 from both a and b. Since c is not at distance n−2 from any of the points of Tz0, we conclude that Tz0 ⊆ Γn−2(c0), a contradiction to (3). If p = m, but if a0 6= projmc 6= b0, we obtain a contradiction considering the line z0 at distance n/2−3 from m on [c, m] that does not contain a0 orb0). So the path [c, l0] contains a0 orb0 (henceδ(c, m) = n/2 + 4). Suppose [c, l0] contains a0. Consider now the element q defined by [m, c]∩[m, a] = [m, q]. Then we first show that q coincides either with a0 (Case 2 below), or with the element a00 = proja0a (Case 1 below). Indeed, if not, then δ(a, c)< n, which implies that δ(a, c) = 4 (by Condition (iv)) and x1 = a1c. Since (b, x1) ∈ O, δ(b, x1) is then equal to n. So it would be possible to find an element of Ta,b for which the projection onto ax1 is different from a and from x1, a contradiction (such a point would be at distance n−2 from x1, which would imply that x1 6∈Sa,b). One checks that in the case n= 10, we end up with the same possibilities.

Case 1. Consider the element m0 ∈[a00, c] that is at distance 2 from a00. A point of Sa,c lies at distance n/2 or n/2 + 2 from m0. Because of the conditions, x1 ∈Sa,c. If x1 lies at distance n/2 + 1 from a0, then δ(x1, m0) =n/2 + 4, a contradiction. If x1 lies at distance n/2 + 1 from b0, there arises a path of length n/2 + 6 between x1 and m0, which is again a contradiction, since n >8. Note that x1 cannot lie at distance n/2 from m because (x1, y)6∈O for y∈Ta,b.

Case 2. Suppose x1 lies at distance n/2 + 1 from b0. Let b0 be the projection of x1 onto b0. Then a point of Sx1,b lies at distance n/2 or n/2 + 2 from b0. Because of the conditions, c ∈ Sx1,b. But we have a path of length n/2 + 6 between c and b0 (containing [c, a0]), a contradiction since n 6= 8. Note that again, δ(x1, m) 6= n/2.

So we know that x1 lies at distance n/2 + 1 from a0. Suppose the projections of c and x1 onto a0 are not equal (which only occurs if n ≡ 2 mod 4, since s = 2). Let a0 = proja0x1. Since c∈Sx1,a, the distance betweencanda0 is eithern/2 orn/2 + 2,

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a contradiction (δ(c, a0) = n/2 + 4). So the projection of c onto a0 is the element a0. Suppose proja0c6= proja0x1. Since the distance betweencand a0 isn/2 + 2, and c∈Sa,x1, the pointc has to lie at distance n/2 + 1 from either a0 or proja0x1, which is not true. So proja0c= proja0x1 :=h. Note that the projections of cand x1 onto h are certainly different, since we know that the distance between c and x1 is either n or 2, and the last choice would contradict the fact that a ∈ Sx1,c. Now consider the projection m0 of c onto h. This is an element at distance n/2 from both c and x1. Now δ(b, m0) = n/2 + 4, which contradicts the fact thatb ∈Sc,x1.

This completes the case s= 2 and hence the casei=n−2.

2.6. Case i = n−1

We can obviously assume n ≥ 6. If n = 6 and s = t = 2, then an easy counting argument yields the result. Ifs= 2, then, since both sand tare infinite forn odd,n is even and hence t > 2. In this case, we dualize the arguments below (this is possible since i is odd). So we may assume throughout thats >2.

For two points a and b with δ(a, b) 6=n−1, let Oa,b be the set of pairs of points {c, c0}, cand c0 different from a and from b, for which

Tv,v0 ⊆Γn−1(w)∪Γn−1(w0),

whenever {a, b, c, c0}={v, v0, w, w0}. For a pair {c, c0} ∈Oa,b, we claim the following:

(i) If δ(a, b) = 2, then either c and c0 are different points on the line ab (distinct from a and b), or, without loss of generality, c is a point onab and c0 ∈Γ3(ab) with projabc0 6∈

{a, b, c}. Moreover, all the pairs (c, c0) obtained in this way are elements of Oa,b. (ii) If δ(a, b) = 4, then either c and c0 are collinear points on the lines am or bm (where

m=a1b) different from m, or cand c0 are points collinear with m, at distance 4 from botha andb, and at distance 4 from each other. Again, all the pairs (c, c0) obtained in this way, are elements of Oa,b.

(iii) Let 4 < δ(a, b) = k < n−1 and put m = a1b. Then c and c0 are points at distance k/2 from m, at distancek from both a and b, and at distancek from each other.

Ifδ(a, b) = 2, then an elementx ofTa,bis either opposite the line ab, or lies at distance n−3 from a unique point on ab, different from a and from b. If δ(a, b) = 4, then an element x of Ta,b either lies at distance n−1 or n−3 from m =a1b with projma 6= projmx6= projmb or lies at distance n−3 from a point x0 on am orbm, x0 6∈ {a, b, m} with am6= projx0x6=bm.

It is now easy to see that the given possibilities forc and c0 in (i) and (ii) indeed satisfy the claim for δ(a, b) = 2 and δ(a, b) = 4, respectively.

Letδ(a, b) = k ≤n−2 and again putm =a1b. Suppose {c, c0} ∈Oa,b. For an element y with δ(m, y) =j ≤n−k/2−2 and projma 6= projmy6= projmb, we define the following set:

Ty ={x∈ P |δ(x, y) = (n±1)−j−k/2 and projyx6= projym if δ(x, y)6=n}.

For an element y with δ(m, y) = n−k/2− 1 and projma 6= projmy 6= projmb, we define Ty as the set of elements at distance 2 from y, not incident with projym. For an element y

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with δ(m, y) = n−k/2 and projma 6= projmy 6= projmb, we define Ty as the set of elements incident with y, different from projym. Note that Ty ⊆Ta,b.

First we make the following observation. Let y be an element for which the set Ty is defined, and for which δ(m, y)≤n−k/2−2. Then there exists an element v ∈ {c, c0} such that Ty ⊆Γn−1(v) if and only if δ(v, y) = δ(a, y) and projyv = projya or projyv = projyb.

Now we prove claims (i), (ii) and (iii) above by induction on the distance k between a and b. Let k≥2. In the sequel, we include the proof for the case k = 2 in the general case.

Suppose first there exists an element v ∈ {c, c0} such that Tm ⊆ Γn−1(v). Then, by the previous observation, δ(v, m) = δ(m, a) = k/2 and we may assume that projmv = projma.

This implies that δ(a, v) ≤ k − 2. Put {c, c0} = {v, v0}. If k = 2, we obtain a = v, a contradiction. If k = 4, then v is a point on the line am, v 6= m, and the only remaining possibility, considering the induction hypothesis and the condition Ta,v ⊆Γn−1(b)∪Γn−1(v0) is thatv0 is also a point onam, different from m. Ifk > 4, the position ofbcontradicts again the fact thatTa,v ⊆Γn−1(b)∪Γn−1(v0) and the induction hypothesis. Indeed, the element at distance δ(a, v)/2 from both a and v does not lie at distance δ(a, v)/2 from b. In this way, we described all the possibilities for the points c and c0 in case there is a point v ∈ {c, c0} for which Tm ⊆ Γn−1(c). So from now on, we assume that there does not exist an element v ∈ {c, c0} such that Tm ⊆Γn−1(v).

Let l be any element incident with m, different from the projection of a or b onto m.

Suppose there exists a point v ∈ {c, c0} such that Tl ⊆ Γn−1(v). Then δ(v, l) = δ(l, a) = k/2 + 1 and we can assume that projlv = projla = m. Since Tm 6⊆ Γn−1(v), we also know that projma6= projmv =:w6= projmb. Put {v, v0}={c, c0}.

Suppose firstk = 2. Thenv is a point on the lineab. We now show that the point v0 lies at distance 2 or 4 fromv such that projvv0 =m. Indeed, suppose projvv0 6=m orδ(v, v0) =n.

If δ(v, v0) 6=n, put γ0 = [v, v0]. If δ(v, v0) =n, let γ0 be an arbitrary n-path between v and v0 not containing m. Let x be an element of Ta,b at distance n−3 from v such that either x lies on γ0, or [v, x] contains γ0. Then x is an element ofTa,b not at distance n−1 from v or v0, a contradiction. So we can assume that projvv0 =m. Suppose now 4 < δ(v, v0). Let Σ be an arbitrary apartment through v and v0. Then the unique element of Σ at distance n−3 from v and belonging to Ta,b, does not lie at distance n−1 from v0, a contradiction, so the distance between v and v0 is 2 or 4. Suppose δ(v, v0) = 4 and projabv0 =b. Then we obtain a contradiction (with the induction hypothesis) interchanging the roles of b and v. So v0 is a point onab, orv0 is a point at distance 3 from abfor which the projection ontoabis different froma,b orv, as claimed in (i).

Suppose now k 6= 2. Let w0 = projwv. Since the distance between v and any element of Tw0 is less than or equal to n−3, we have that Tw0 ⊆ Γn−1(v0), from which follows that δ(v0, w0) = δ(w0, a) = k/2 + 2 and projw0v0 = projw0a = w. Since Tm 6⊆ Γn−1(v0), we either have that v0 is a point at distance k/2 from m for which the projection onto m is different from w and projma 6= projmv0 6= projmb (as required in (ii) and (iii)), or v0 is a point at distance k/2 + 2 from m for which the projection onto m is w. In the latter case, let z be the projection of v0 onto w (then δ(v, z) = δ(v0, z) = k/2) and consider an element x at distance n−1−k/2−2 from z such that projzv 6= projzx 6= projzv0. Then x is an element of Ta,b at distance n−3 from both v and v0, a contradiction. In this way, we described all the possibilities for the points c and c0 in case there is a point v ∈ {c, c0} and an element l

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as above for which Tl ⊆ Γn−1(c). So from now on, we assume that there does not exist an element v ∈ {c, c0} such that Tl⊆Γn−1(v), for any l as above.

We now prove that

(3) if yis an element for which the set Ty is defined, withδ(m, y)>1, then there does not exist a point v ∈ {c, c0} such that Ty ⊆Γn−1(v).

This is done by induction on the distance j between y and m.

So let by way of contradiction l be an element at distance j from m, j > 1, for which the set Tl is defined and such that there exists an element v ∈ {c, c0} with Tl ⊆ Γn−1(v).

Put {v, v0} = {c, c0}. Let first j < n −k/2−1. Then δ(v, l) = δ(l, a) = k/2 +j and w := projlv = projla but u := projwv 6= projwa. Let w0 = projuv. Note that the distance between w0 and an element of Tw0 is (n ± 1)− k/2 −(j + 1), so an element of Tw0 lies at distance at most n −3 from v. We conclude that Tw0 ⊆ Γn−1(v0), from which follows that δ(v0, w0) = δ(a, w0) = k/2 +j + 1 or δ(v0, w0) = n− 3 (the latter is possible only if j =n−k/2−2), and projw0v0 = projw0a=u. Let projwa=u0. First supposeδ(v0, w0)6=n−3.

From the assumptions, it follows that projwv0 6=u0. Depending on whether the projection of v0 onto w is u or not, the distance between v0 and u0 is k/2 +j + 2 or k/2 +j. Note that δ(v, u0) = k/2 +j. Now consider an element x at distance (n−1)−(k/2 +j) from u0 such that proju0x6=w, and such that x either lies on [u0, b], or [u0, x] contains [u0, b]. Then x is an element of Tv,v0 not contained in Γn−1(a)∪Γn−1(b), a contradiction. Ifδ(v0, w0) =n−3, then we similarly obtain a contradiction. So Ty 6⊆ Γn−1(v), for any element y at distance j from m.

Now let j = n −k/2−1. Note that Tl consists of all elements at distance 2 from l, not incident with l0 = projlm. Then δ(v, l) = n−1 or δ(v, l) = n−3, and in both cases, projlv = projla. If δ(v, l) = n−1 (= δ(a, l)), we proceed as in the previous paragraph and end up with a contradiction. So let δ(v, l) =n−3. First suppose that projl0v 6= projl0a=w.

Now consider an element w0 incident with w,l0 6=w0 6= projwa and w0 6= projwb. ThenTw0 ⊆ Γn−1(v), a contradiction sinceδ(m, w0) =j −1. So projl0v =w. Let [u, m] = [v, m]∩[w, m]

and put u0 = projuv. Suppose first that projma6=u0 6= projmb and v 6=m (v =m can occur only if k = 4). Then Tu0 ⊆ Γn−1(v0). Indeed, if we put i= δ(u, l), then δ(v, u0) = n−4−i andδ(m, u0) = n−k/2−i. So the distance betweenu0 and an element ofTu0 isi±1, and the distance between v and an element of Tu0 is at most n−3. So Tu0 is contained in Γn−1(v0), which is a contradiction since δ(m, u0) < j (indeed, i ≥ 2). Suppose finally u0 = projma or v =m. If k= 2, we end up with a pointv lying on ab(namelyv = projml). But then, for an arbitrary pointxonm, different froma,b andv, we have thatTx ⊆Γn−1(v), in contradiction with our assumptions. Ifk= 4, we end up with v =m, but then the position ofbcontradicts the fact that Ta,v ⊆ Γn−1(b)∪Γn−1(v0) and the (general) induction hypothesis. Finally, if k >4, then δ(v, a)≤δ(v,projma) +δ(a,projma) =k−4. Now the position of b contradicts again the fact that Ta,v ⊆Γn−1(b)∪Γn−1(v0) and the (general) induction hypothesis.

Let finally j = n−k/2. Note that Tl consists of all elements incident with l, different from the projection l0 of m onto l. Then δ(v, l0) = n−1 or δ(v, l0) = n−3. Note that, in both cases, projl0v 6= projl0a. Indeed, projl0v = projl0a would imply that Tl0 ⊆ Γn−1(v), a contradiction with our assumptions. Suppose firstδ(v, l0) = n−3. Letl00= projl0v. Since no element incident with l00 is at distance n−1 from v, we have Tl00 ⊆Γn−1(v0), which implies

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thatδ(v0, l0) is eithern−3 orn−1 and projl0v0 6= projl0a. Consider now the element on [a, l0] at distance 2 from l0. This is an element of Tv,v0 which is at distance n−3 from botha and b, a contradiction. Suppose now δ(v, l0) = n−1. Let x be the element on [v, l0] at distance 2 from l0. Since xis the only element of Tl0 not at distance n−1 from v, this elementx lies at distance n−1 from v0. But then δ(v0, l0) is either n−1 or n−3. If projl0v0 = projl0a, then Tl0 ⊆ Γn−1(v0), a contradiction with our assumptions. If projl0v0 6= projl0a, then again the element on [a, l0] at distance 2 froml0 is an element of Tv,v0 at distance n−3 from both a and b, the final contradiction. This proves (3).

Suppose nowTy 6⊆Γn−1(v) for allv ∈ {c, c0}and for any appropriate elementy. Consider an elementlat distancen−k/2 frommsuch that the projection ofl ontomis different from the projections ofaand bontom. Letube the projection ofaontol. SinceTl 6⊆Γn−1(c) and Tl 6⊆Γn−1(c0), there is an elementxincident with l, different fromu, at distancen−1 fromc but not fromc0, and an elementy incident withl, different from u, at distance n−1 from c0 but not from c. Soδ(x, c0) =n−3 =δ(y, c) and projxc0 6=l 6= projyc. But from this follows that, for an arbitrary element l0 incident with u, l 6= l0 6= projua, we have Tl0 ⊆ Γn−1(c), a contradiction. This proves the claims (i), (ii) and (iii).

For two points a, b, let Ca,b be the set containing a, b, and all points c for which there exists a point c0 such that {c, c0} ∈ Oa,b. Now let S be the set of pairs of points (a, b), δ(a, b)6=n−1, for which there does not exist an element at distancen−1 from all the points of Ca,b. We claim that S contains exactly the pairs of points (a, b) for which δ(a, b) = 2 or δ(a, b) = 4.

First assumeδ(a, b) = 2. Let by way of contradiction w be an element at distance n−1 from all points ofCa,b. Since all the points of the lineabare contained inCa,b,wlies opposite ab. If v is an arbitrary point on ab, different from a and from b, then the element on [v, w]

that is collinear withv, is contained inCa,b, but lies at distancen−3 fromw, a contradiction.

Suppose nowδ(a, b) = 4. Let by way of contradictionwbe an element at distancen−1 from all points ofCa,b. Then w lies at distance n−1 from all the points collinear with m=a1b, which is not possible. Finally suppose 4< δ(a, b) =k 6=n−1. Let a0 be the element on the path [a, b] at distance k/2−1 froma, and xan element at distance (n−1)−(k/2−1) from w with proja0a 6= proja0w 6= proja0b. Then w lies at distance n−1 from all points of Ca,b. Our claim is proved.

Now let S0 be the subset ofS containing all the pairs (a, b) with the property that there exist points x and x0 belonging to Ca,b such that (x, x0) 6∈ S. Then S0 contains exactly the pairs of collinear points. Indeed, if δ(a, b) = 2, we can find pointsxand x0 inCa,bat distance 6 from each other, while if δ(a, b) = 4, then Ca,b ⊆ Γ2(m). This completes the proof of the case i=n−1.

2.7. Case i = n/2

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